Continuing with Chisigma's solution, then I'll finish it if that's alright. I'll use variation of parameter:

Then and therefore

Let

Substituting that into the original DE we obtain the following two equations to solve:

Then

Substituting that one into the second one, I get or so that and in which we drop the arbitrary constants in this case since we're only looking for a particular solution to add to the complimentary solution.